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4.2. Construction of the second trial function

4.3. Symmetric vs asymmetric potential

5. The N-dimensional problem

Appendix A. A soluble example

A.1. A two-level model

A.2. Square-well example (Cont.)

A.3. The iterative sequence

References

1. Introduction

Quantum physics is largely governed by the Schroedinger equation. Yet, exact solutions of the equation are relatively few. Besides lattice and other numerical calculations, we rely mostly on perturbative expansions. Such expansion quite often leads to a divergent series with zero radius of convergence, as in quantum electrodynamics, quantum chromodynamics, and problems involving tunneling and instantons. In a series of previous papers [1], [2], [3] and [4] we have presented a new approach to solve the low lying states of the Schroedinger equation. In the special case of one-dimensional problems, this new approach leads to explicit convergent iterative solutions, in contrast to perturbative series expansions. These convergent solutions include the long standing difficult problem [5], [6], [7], [8], [9], [10], [11], [12], [13] and [14] of a quartic potential with symmetric minima.

In this paper, we discuss some additional results bearing on the new method. In the one-dimensional case, we show that by changing the boundary condition to be applied at each iteration, we can obtain a convergent alternating sequence for the ground state energy and wave function instead of the monotonic sequence found before [4]. This result will be spelled out later in this section and proved in Section 3. We also find that the asymmetric quartic double-well potential can be treated by an extension of the procedure used previously for the symmetric case. This extension is treated in Section 4.

In addition, we have begun the exploration of higher dimensional problems along the same line. Although the same kind of iterative procedure can be set up, the linear inhomogeneous equation to be solved at each step cannot now be reduced to simple quadratures, as was done for one dimension. However, it is of interest that this equation is identical in form to an electrostatic analog problem with a given position-dependent dielectric constant media; at each nth iteration, there is an external electrostatic charge distribution determined by the (n − 1)th iterated solution, as we shall discuss in this section.

Consider the Schroedinger equation

Hψ=Eψ,

(1.1)

where H is the Hamiltonian operator, ψ the wave function, and E its energy. For different physics problems, H assumes different forms. For example, for a system of n non-relativistic particles in three dimensions, H may be written as

(1.2)

where x stands for x₁, x₂, … , x_3n the coordinate components of these n particles, V (x) is the potential function, C_ij are constants, and p₁, p₂, … , p_3n are the momentum operators satisfying the commutation relation

[pi,xj]=-iδij.

(1.3)

(Throughout the paper, we set Planck’s constant  = 1.) For a relativistic field theory, the Hamiltonian usually takes on a different form. Let Φ (r) be a scalar boson field at a three-dimensional position vector r, and Π (r) be the corresponding conjugate momentum operator. In this case we may write

H=∫d3r[Π2(r)+V(Φ(r))]

(1.4)

with Π (r) and Φ (r′) satisfying the commutation relation

[Π(r),Φ(r′)]=-iδ3(r-r′).

(1.5)

In both cases, the dependence of H on the momentum operators p_i and Π (r) are quadratic. Consequently, they can be brought into an identical standard form. In the above case of a system of non-relativistic particles, through a linear transformation

{xi}→{qi},

(1.6)

the Hamiltonian (1.2) can be written in the standard form

(1.7)

with

(1.8)

Likewise for the relativistic boson field Hamiltonian (1.4), we can use the Fourier-components of Φ (r) and Π (r) as the set {x_i} and {p_i}. Through a similar transformation (1.6), the field Hamiltonian (1.4) can also be brought into the standard form Figs. (1.7) and (1.8), but with the number of variables N = ∞. All our subsequent discussions will start from the Schroedinger equation in this standard form Figs. (1.7) and (1.8). Furthermore, in this paper, we shall limit our discussions only to the ground state. In order to solve

(1.9)

where q stands for the set {q_i}, we proceed as follows:

1. Construct a good trial function  (q). A rather efficient way to find such trial functions is given in the next section.

2. By differentiating , we define

(1.10)

in which the constant E₀ may be determined by, e.g., setting the minimum value of U (q) to be zero. Thus,  (q) satisfies a different Schroedinger equation

(1.11)

Define w (q) and by

U(q)=V(q)+w(q)

(1.12)

and

(1.13)

The original Schroedinger equation (1.9) can then be written as

(1.14)

Multiplying this equation on the left by  (q) and (1.11) by ψ (q), we find their difference to be

(1.15)

The integration of its left-hand side over all space is zero, which yields

(1.16)

3. The above equation (1.14) will be solved iteratively by considering the sequences

(1.17)

that satisfy

(1.18)

with

ψ0(q)= (q).

(1.19)

As in Figs. (1.15) and (1.16), we multiply (1.11) by ψ_n and (1.18) by ; their difference gives

(1.20)

and therefore

(1.21)

As we shall show, for many interesting problems

(1.22)

in contrast to the perturbative series expansion using w (q) as the perturbation. The key difference lies in the above expression (1.21) of , which is a ratio, with both its numerator and denominator depending on the (n − 1)th iterative solution ψ_n − 1.

4. There exists a simple electrostatic analog problem for the iterative equation (1.18). Assuming that ψ_n − 1 (q) has already been solved, we can determine through (1.21). The right-hand side of (1.20), defined by

(1.23)

is then a known function. Introduce

fn(q)≡ψn(q)/ (q).

(1.24)

In terms of f_n (q), the nth order iterative equation (1.20) becomes

(1.25)

Consider a dielectric medium with a dielectric constant dependent on q, given by

κ(q)≡ 2(q).

(1.26)

Interpret σ_n (q) as the external electrostatic charge distribution, the electrostatic potential, the electrostatic field, and

(1.27)

the corresponding displacement vector field. Thus (1.25) becomes

·Dn=σn,

(1.28)

the Maxwell equation for this electrostatic analog problem.

At infinity,  (∞) = 0. In accordance with Figs. (1.26) and (1.27), we also have D_n (∞) = 0. Hence the integration of (1.28) leads to the total external electrostatic charge to be also zero; i.e.,

(1.29)

which is the same result given by (1.21) for the determination of . Because the dielectric constant κ (q) in this analog problem is zero at q = ∞, the dielectric media becomes a perfect dia-electric at ∞. Thus, the equation of zero total charge, given by (1.29), may serve as a much simplified model of charge confinement, analogous to color confinement in quantum chromodynamics.

We note that (1.25) can also be derived from a minimal principle by defining

(1.30)

Because of (1.29), the functional I (f_n (q)) is invariant under

fn(q)→fn(q)+constant.

(1.31)

Since the quadratic part of I (f_n (q)) is the integral of the positive definite , the curvature of I (f_n (q)) in the functional space f_n (q) is always positive. Hence, I (f_n (q)) has a minimum, and that minimum determines a unique electrostatic field , as we shall see. To establish the uniqueness, let us assume two different f_n, both satisfy (1.25), with the same κ =  ² and the same σ_n; their difference would then satisfy (1.25) with a zero external charge distribution. For σ_n = 0, the minimum of I (f_n (q)) is clearly zero with the corresponding f_n = 0. To derive f_n (q) from f_n, there remains an additive constant at each iteration. As we shall show, this arbitrariness allows us the freedom to derive different types of convergent series.

To illustrate this freedom, let us consider a one-dimensional problem in which we may replace the variables {q_i} by a single x. Furthermore, for this discussion, let us assume the potential V (x) to be an even function, with

V(x)=V(-x)

(1.32)

(a condition that will be relaxed in our later analysis). The evenness of V (x) requires ψ (x) = ψ (−x) and therefore also  (x) =   (−x). Thus, we need only to consider the half-space

x 0.

(1.33)

Eqs. Figs. (1.24), (1.27) and (1.28) can be written now as

fn(x)=ψn(x)/ (x),

(1.34)

(1.35)

and

(1.36)

where

κ(x)= 2(x)

(1.37)

and

(1.38)

same as before. Throughout the paper, ′ denote .